Final answer:
The probability of finding 3 or fewer errors in 100 pages of a textbook with a mean of 0.01 errors per page is 0.9810 when calculated using the Poisson distribution with the mean number of errors for 100 pages (λ = 1).
Step-by-step explanation:
The question involves calculating the probability of an event using the Poisson distribution, specifically the probability of finding 3 or fewer errors in 100 pages of a textbook when the mean number of errors per page is 0.01. First, since the mean is per page, we multiply the given mean by the number of pages to find the mean number of errors in 100 pages: λ = 0.01 errors/page × 100 pages = 1 error.
Using the Poisson distribution function for λ = 1, we want to find P(X ≤ 3), which is the sum of probabilities of x = 0, 1, 2, and 3, where X represents the number of errors in 100 pages. We use the Poisson probability mass function (PMF) for each of these values:
- P(X=0) = e^(-1) × 1^0 / 0! = 0.3679
- P(X=1) = e^(-1) × 1^1 / 1! = 0.3679
- P(X=2) = e^(-1) × 1^2 / 2! = 0.1839
- P(X=3) = e^(-1) × 1^3 / 3! = 0.0613
Adding these up, we get P(X ≤ 3) = 0.3679 + 0.3679 + 0.1839 + 0.0613 = 0.9810. Therefore, the probability that there are 3 or less errors in 100 pages is 0.9810, when rounded to four decimal places.